= Solution
In an <Einstein-de Sitter universe>, take $a(t)=(t/t_0)^{2/3}$ and $t_0=2/(3H_0)$. The galaxy seen today at $z=1$ emitted the light now reaching us when $a(t_e)=1/2$, so $t_e/t_0=2^{-3/2}$. Its fixed <comoving radial distance> is
$$
\chi=c\int_{t_e}^{t_0}\frac{dt}{a(t)}
=3ct_0\left(1-\frac1{\sqrt2}\right).
$$
Put $b=1-1/\sqrt2$. A new message sent now reaches the galaxy at $t_1$, where
$$
\chi=3ct_0[(t_1/t_0)^{1/3}-1],\qquad
\frac{t_1}{t_0}=(1+b)^3.
$$
Assume an immediate reply and negligible local handling delays. The return consumes the same additional <conformal time>, giving the <Einstein-de Sitter round-trip message time>
$$
\boxed{\frac{t_2}{t_0}=(1+2b)^3=(3-\sqrt2)^3\simeq3.98781,\qquad
\Delta t=t_2-t_0\simeq2.98781t_0=\frac{1.99187}{H_0}.}
$$
For the numerical value $H_0=72\,\mathrm{km\,s^{-1}\,Mpc^{-1}}$ supplied later in the paper, $H_0^{-1}=13.58\,\mathrm{Gyr}$ and the waiting time is about $27.1\,\mathrm{Gyr}$. The age $t_0=9.05\,\mathrm{Gyr}$ here belongs to the stipulated matter-only model, not a claim about the actual measured cosmic age.
After $n$ one-way legs, $t_n=t_0(1+nb)^3$. Since $\int_{t_0}^{\infty}dt/a(t)=\infty$, every finite number of messages is possible and there is no finite fundamental upper count. The exchange takes arbitrarily long, with successively longer intervals; infinitely many messages are not completed in finite cosmic time.
\Image[/past-exam-of-the-general-relativity-and-cosmology-course-of-the-university-of-oxford/2018-cosmological-message-times.png]
{title=Message exchange depends on the remaining conformal time, not just the present expansion rate}
{description=The left conformal spacetime diagram shows repeated light-signal legs between two comoving galaxies in an Einstein-de Sitter model. The right diagram illustrates a finite future conformal interval: one outward leg fits but a complete round trip does not. The right interval is a schematic example, not a parameter estimate for our Universe.}
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